1/3 + 1/6 + 1/10 + ... + 2/n(n+1) = 2003/2004
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4008-1=4007\)